Answer:
Given, Tangents AM and AN are drawn from point A to a circle with centre O. To prove: \[AM=AQ\] Construction: Join OM, ON and OA Proof: Since AM is a tangent at M and OM is radius \[\therefore \] \[OM\bot AM\] Similarly, \[ON\bot AN\] Now, in \[\Delta \,OMA\] and \[\Delta \,ONA\]. \[OM=ON\] (radii of same circle) \[OA=OA\] (common) \[\angle OMA=\angle ONA=90{}^\circ \] \[\therefore \Delta \,OMA\cong \Delta \,ONA\] (By RHS congruence) Hence, \[AM=AN\] (By cpct) Hence Proved.
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